Question

A sphere of radius $0.1m$ and mass $8 \pi \text{ kg}$ is attached to the lower end of a steel wire of length $5.0m$ & diameter $10^{- 3}m$ . The wire is suspended from $5.22m$ high ceiling of a room. When the sphere is made to swing like a simple pendulum, it just grazes the floor at its lowest point. Calculate the velocity of the sphere at the lowest position. $Y$ for steel = $1.994\times 10^{11}Nm^{- 2}$

• A. $7.7ms^{- 1}$
• B. $4.4ms^{- 1}$
• C. $2.2ms^{- 1}$
• D. $8.8ms^{- 1}$

Answer 1

Answer: (D)

As the length of the wire is $5m$ and diameter $2\times 0.1=0.2m$ and at the lowest point, it grazes the floor which is at a distance $5.22m$ from the roof, the increase in the length of the wire at lowest point

$\Delta \text{L} = 5 \text{.} 2 2 - \left(5 + 0 \text{.} 2\right)$
$= 0 \text{.} 0 2 \text{ m}$
So the tension in the wire (due to elasticity)
$T=\frac{Y A}{L}\Delta L=\frac{1 \text{.} 9 9 4 \times 1 0^{1 1} \times \pi \left(5 \times 1 0^{- 4}\right)^{2} \times 0 \text{.} 0 2}{5}=199\text{.}4\pi \text{N}$
and as the equation of circular - motion of a mass $m$ tied to a string in a vertical plane is
$\frac{m v^{2}}{r}=T-m\text{g cos}\theta $
So at the lowest point
$\frac{m v^{2}}{r}=T-m\text{g}$ $\left[\text{as } \theta = 0\right]$
But here $r=5+0.02+0.1=5.12m$
So $\left(\frac{8 \pi v^{2}}{5 . 12}\right)=199\text{.}4\pi -8\pi \times 9\text{.}8$
i.e., $\left(\text{v}\right)^{2} = \left(1 2 1 \times 5 \text{.} 1 2 / 8\right) = 7 7 \text{.} 4 4$
So $v=8.8ms^{- 1}$